## Characterization of those semigroups with zero on which every equivalence relation is a congruence

Let $S$ be a semigroup with zero. Show that if every equivalence relation on $S$ is a congruence, then $S$ is either a zero semigroup or a semilattice of order 2.

Let $S$ be a semigroup with a zero element $0$, and suppose that every equivalence relation on $S$ is a congruence.

Suppose $a,b \in S$ such that $ab \neq 0$. Consider now the equivalence relation $\sigma$ on $S$ whose classes are $\{\{a,0\}, S \setminus \{a,0\}\}$. By our hypothesis, $\sigma$ is a congruence. Now $a \sigma 0$, so $ab \sigma 0b = 0$. Since $ab \neq 0$, we have $ab = a$. Now consider the equivalence $\tau$ on $S$ whose classes are $\{\{b,0\}, S \setminus \{b,0\}\}$. Again $\tau$ is a congruence, and we have $b \tau 0$, so that $ab \tau a0 = 0$. Since $ab \neq 0$, we have $ab = b$. So $a = b$. Moreover, $a^2 = a$.

Thus we have $S = Z \cup E$, where $Z = \{s \in S \ |\ sS = 0\}$ and $E = S \setminus Z$. By the preceding argument, in fact $E = \{s \in S\ |\ s^2 = s, s \neq 0\}$ is the set of nonzero idempotents in $S$.

We claim that $E$ contains at most one element. To see this, suppose $E \neq \empty$ and let $e,f \in E$. Consider the equivalence $\sigma$ on $S$ whose classes are $\{\{e,f\},S \setminus \{e,f\}\}$. Now $e \sigma f$, so that $e = ee \sigma ef$. So either $ef = e$ or $ef = f$. If $ef = e$, consider the equivalence $\tau$ whose classes are $\{\{f,0\}, S\setminus\{f,0\}\}$. Now $f \tau 0$, so $ef \tau 0$. Since $ef = e \neq 0$, $ef = f$, and thus $e = f$. If $ef = f$, consider the equivalence $\tau$ whose classes are $\{\{e,0\}, S \setminus \{e,0\}\}$. By a similar argument, $e = f$. So $E$ contains at most one element.

If $E = \emptyset$, then $S = Z$ is a zero semigroup.

Suppose instead that $E = \{e\}$, and let $z \in Z$. Consider the equivalence $\sigma$ whose classes are $\{\{e,z\}, S \setminus \{e,z\}\}$. Now $e \sigma z$, so that $e = e^2 \sigma ez = 0$. So $e \sigma 0$, and since $e \neq 0$, $z = 0$. So $S = \{e,0\}$. Certainly $S$ is a semilattice of order 2.